Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 229

Particle A is projected vertically upward from a top of a tower. At the same time particle \(B\) is dropped from the same point. The graph of distance (s) between the two particle varies with time is.

Short Answer

Expert verified
The distance (s) between particles A and B varies linearly with time (t), and can be represented by the equation \(s = ut\), where u is the initial velocity of particle A. The graph for this relation is a straight line with a positive slope (u) and has the distance s on the vertical axis and time t on the horizontal axis.

Step by step solution

01

Write the equations of motion for particles A and B

Since particle A is projected vertically upward, its initial velocity will be a non-zero value. We'll denote the initial velocity of A as \(u\). Particle B, on the other hand, is dropped from rest, meaning its initial velocity is equal to 0. Let's denote the acceleration due to gravity as \(g\) (a negative value). We can then write the equations of motion for particles A and B using the second equation of motion: For particle A: \(s_A = ut - \frac{1}{2}gt^2\) For particle B: \(s_B = -\frac{1}{2}gt^2\)
02

Find the distance between particles A and B

To find the distance \(s\) between the particles, we'll subtract particle B's displacement from particle A's displacement: \(s = s_A - s_B = (ut - \frac{1}{2}gt^2) - (-\frac{1}{2}gt^2) = ut - \frac{1}{2}gt^2 + \frac{1}{2}gt^2\) Simplifying the equation, we get: \(s = ut\)
03

Plot the graph of distance between particles A and B

The equation we derived in step 2 is a linear equation of the form \(s = ut\), where \(s\) is the distance between the particles, and \(t\) is the time. This means that the graph of distance between the particles A and B versus time will be a straight line with a positive slope (u). To plot the graph, we will have the distance s on the vertical axis and time t on the horizontal axis. At the initial state (t=0), the distance between the particles will be 0, which will be the starting point of the graph. As time progresses, the distance between the particles will increase linearly, and the graph will have a positive slope equal to the initial velocity of particle A (u).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle is moving in a straight line with initial velocity of $200 \mathrm{~ms}^{-1}\( acceleration of the particle is given by \)\mathrm{a}=3 \mathrm{t}^{2}-2 \mathrm{t}$. Find velocity of the particle at 10 second. (A) \(1100 \mathrm{~ms}^{-1}\) (B) \(300 \mathrm{~ms}^{-1}\) (C) \(900 \mathrm{~ms}^{-1}\) (D) \(100 \mathrm{~ms}^{-1}\)

A particle goes from point \(\mathrm{A}\) to \(\mathrm{B}\). Its displacement is \(\mathrm{X}\) and path length is \(\mathrm{y}\). So $\mathrm{x} / \mathrm{y} \ldots$ \((\mathrm{A})>1\) (B) \(<1\) (C) \(\geq 1\) (D) \(\leq 1\)

Train \(A\) is \(56 \mathrm{~m}\) long and train \(\mathrm{B} 54 \mathrm{~m}\) long. They are travelling in opposite direction with velocity $15(\mathrm{~m} / \mathrm{s})\( and \)5(\mathrm{~m} / \mathrm{s})$ respectively. The time of crossing is. (A) \(12 \mathrm{~s}\) (B) \(6 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(18 \mathrm{~s}\)

\(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are equal vectors what from the followings is true. (A) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are anti parallel (B) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are parallel (C) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) may be perpendicular (D) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{\rightarrow}\) may be free vectors

Which from the following is a scalar? (A) Electric current (B) Velocity (C) acceleration (D) Electric field
See all solutions

Recommended explanations on English Textbooks

Listening and Speaking

Read Explanation

Summary Text

Read Explanation

Argumentative Essay

Read Explanation

Cues and Conventions

Read Explanation

Essay Plan

Read Explanation

Pragmatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free